package case3; public class superTest { public static void main(String[] args) { // 实例化一个Student类的对象s，为Student对象s中的school赋值，打印输出信息 /********* begin *********/ Student s = new Student("张三",18,"哈佛大学"); s.print(); /********* end *********/ } } class Person { /********* begin *********/ String name; int age; public void Person(String name,int age){ this.name=name; this.age=age; } /********* end *********/ } class Student extends Person { /********* begin *********/ String school; public void Student(String name,int age,String school){ super.Person(name,age); this.school=school; } void print(){ System.out.println("姓名："+name+"年龄："+age+"学校："+school); } /********* end *********/ }src/case3/superTest.java:7: error: constructor Student in class Student cannot be applied to given types; Student s = new Student("张三",18,"哈佛大学"); ^ required: no arguments found: String,int,String reason: actual and formal argument lists differ in length报错这个，请在不要改变我原有程序架构的条件下修改我的程序

请修改以下代码：

public void Person(String name,int age){
    this.name=name;
    this.age=age;
}

修改为：

public Person(String name,int age){
    this.name=name;
    this.age=age;
}

public void Student(String name,int age,String school){
    super.Person(name,age);
    this.school=school;
}

修改为：

public Student(String name,int age,String school){
    super(name,age);
    this.school=school;
}

在Java中，构造函数没有返回值类型，因此在定义构造函数时不需要声明void。在调用超类的构造函数时，应使用super关键字，并传递相应的参数。